6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
We’re almost at the finish line! Charge control is the last major topic that we will cover this term.
Why charge control? Hasn’t the hybrid-π model served us well this whole semester? It certainly has.
But to see some of the holes that it has left, ask yourself the following questions:
•
How does one analytically treat a device that has entered saturation?
•
Are there any dynamics associated with suddenly changing regions of operation?
•
What really happens if we suddenly cut-off the base current of a BJT?
The list goes on, and the point is that the answers to these questions can be useful. Transistors, after
all, aren’t used only for small signal amplification. Sometimes they are useful as switches, either for
realizing digital logic or in digital-to-analog converters.
We should emphasize that charge control, like open circuit time constants, is not good for giving us
precise numerical answers. It does give us design insight, telling us, for instance, what topologies
should yield high switching speeds.
To review, recall that in lecture we looked at finding Ic(t) in the following situation:
IC
I Bu(t)
iC =
qF
τF
iB =
qF dqF
+
τ BF
dt
⎛ 1
1 ⎞ dqF
iE = −qF ⎜ +
−
⎝ τ F τ BF ⎟⎠ dt
↑
Since everything depends on qF , we first figure out qF (t) :
iB =
qF dqF
+
= I Bu(t)
τ BF
dt
6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
Using basic differential equations, we solved this and found:
(
qF = τ BF I B 1 − e−t /τ BF
IC =
)
qF τ BF
=
I B 1 − e−t /τ BF = β I B 1 − e−t /τ BF
τF τF
(
)
(
)
Now, for the class exercise we’re going to solve this a different way.
CLASS EXERCISE: Consider the hybrid-π model
I C (t)
Vπ
I Bu(t)
↑
rπ
Cπ
↓
gmVπ
Derive the collector current as a function of time. You may find the following reminders helpful:
(1)
gm rπ = β F
(2)
⎛β ⎞
rπ Cπ = ⎜ F ⎟ ( gmτ F ) = β Fτ F = τ BF
⎝ gm ⎠
(Workspace)
Page 2
6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
The equivalence is a little surprising, isn’t it? Think about why this worked out…noticing that gm rπ
and rπ Cπ are both independent of collector current.
Very well. We promised design insight as a result of our derivations, and we’re coming to a first
example. According to our charge control model, turning on a transistor is simply a matter of placing
the right amount of charge in the base.
When we tried the following circuit in lecture:
RL
V0
RB
vI (t) = (VI + VBE ) u(t)
→
+
vI (t)
IB
−
There was a limit to how rapidly we could change the amount of charge in the case specifically:
dqF
V
≤ I
dt
RB
But what if we could arrange to put charge into the base very quickly? Consider:
RL
iCB
vI (t) = (VI + VBE ) u(t)
CB
V0
+
vI (t)
RB
−
Page 3
6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
Hmm. Conceptually, the voltage across the capacitor changes instantaneously from zero to VI :
iCB = C B
dVCB
= CbVI δ (t)
dt
Therefore, the base current can be written:
iB (t) =
VI
u(t) + C BVI δ (t)
RB
Using charge control, we have
iB =
qF dqF VI
+
=
u(t) + C BVI δ (t)
τ BF
dt
RB
The solution is
qF (t) = τ BF
Notice that if we choose C B =
(
)
VI
1 − e−t /τ BF u(t) + C BVI e−t /τ BF
RB
τ BF
,
RB
qF (t) = τ BF
VI
u(t)
RB
Page 4
6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
iC reaches its final value instantaneously! For this reason, C B is sometimes called a speed-up
capacitor. Step responses for different values of C B would look like the following:
IC
C B too high
C B optimum
C B too low
t
Real circuits do exactly this! C B may require some trimming, but this is a trick that works in the
real world.
Page 5
6.301 Solid-State Circuits
Recitation 24: Charge Control w/Demo
Prof. Joel L. Dawson
Charge control also gives us insight into “emitter switching.” That is,
IC
↓
I E u(t)
Challenge: Find I C and I B . We’ll talk more about this case later.
Page 6
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6.301 Solid-State Circuits
Fall 2010
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